Question

A capacitor $C(=100\mu F)$ is charged with a total charge of $1.2\times {10}^{-3}C$ and connected in a circuit as shown. Find the current in the circuit after the key $k$ is closed $(t=0)$ as a function of time ($t$). Notice that the positively charged plate of the capacitor is connected to the negative terminal of the cell.

Answer 1

Let the current in the circuit be $i\left(t\right)$ and the charge on the capacitor be $q\left(t\right)$. Note that the initial polarity of charge $q_{10}$ on the capacitor is opposite to the final steady state charge
Applying Kirchoff's laws
$\frac{q}{C}+R\left(\frac{dq}{dt}\right)=E;E=12V,R=10\times {10}^3\Omega ,C=100\times {10}^{-6}F$
$or\left(\frac{dq}{dt}\right)+\frac{q}{C}=\frac{E}{R}$
Solving this, $q=C_1{e}^{-t/RC}+CE$
where $C_1$ is an arbitrary constant
we determine $C_1$ from the initial condition, where $q_0=1.2\times {10}^{-3}C$
$q=-q_0;-q_0=C_1+CE$
Thus, $q\left(t\right)=-{(CE+q_0)}_1{e}^{-t/RC}+CE$
The current, $i=\frac{dq}{dt}=\left(\frac{CE+q_0}{RC}\right)e^{-t/RC}$
substituting the values $i=\left(2.4mA\right)e^{-t/1sec}$